This document contains the design calculations to check whether an ISLB 350@ 49.5kg/m laterally supported beam can support a maximum bending moment of 180 kN-m and shear force of 210 kN over a 6m span. The calculations show that the beam section has sufficient design bending and shear strength as well as adequate resistance against web buckling and web crippling. Therefore, the ISLB 350@ 49.5kg/m section satisfies the design requirements for this beam.

1. Sanjivani Rural Education Society's
Sanjivani College of Engineering, Kopargaon 423603.
-Department of Civil Engineering-
By
Mr. Sudhir B. Gayake (Assistant Professor)
Mail Id- gayakesudhircivil@sanjivani.org.in
Prof. Sudhir B. Gayake, SCOE Kopargaon

2. Prof. Sudhir B. Gayake, SCOE Kopargaon
Question- Check whether ISLB 350@ 49.5kg/m laterally supported throughout for a
span of 6m for the following data;
(1)Maximum factored BM= 180kN
(2) Maximum factored SF= 210kN
(3) Steel grade=Fe410.
1) Given Data-
Span=l=6000mm,
Mu=140kN-m
Vu=210kN,
fu=410MPa,
fy=250MPa,
ϒm0=1.1
2) Classification of cross section- (Table No 2)
a) Flange= (bf/tf)=[(bf/2)/(tf)]=(82.5/11.4)= 7.23<9.4ɛ ∴ flange is plastic
b) Web= (h1/tw)= (288.3/7.4) = 38.95< 84ɛ ∴ web is also plastic
Properties of ISLB350 @ 49.5 kg/m
h=350mm, bf=165mm, tf=11.4mm,
tw=7.4mm, Izz=13158.3×104mm4 ,
h1=288.3mm,
Zp=851.11×103 mm3
Ze= 751.9×103 mm3
r1=16mm
Section is
Plastic βb=1

3. Prof. Sudhir B. Gayake, SCOE Kopargaon
3) Design shear strength of section-(Cl.No.8.4.1 Pg.No.59)
Vd=[(fyw×Aw)/(√3×ϒm0)]==[(250×350×7.4)/(√3×1.1)]= 339.84kN> Vu………..(OK)
Check for high or low shear condition= 0.6Vd= 0.6×339.84= 203.90kN < Vu…….(High shear)
4) Design bending strength for high shear condition-(Cl No. 8.2.1.3 Pg.no. 53)
For plastic and compact sections-
Md=Mdv =Md – β(Md –Mfd ) ≤ (1.2Zefy/ϒm0)
Mfd= (zfd×fy/γm0)
=[Af (h-tf)]×fy/γm0
=[ 165×11.4 (350-11.4)] ×(250/1.1)
= 144.75 kN-m
Md= (βb×Zp×fy/ϒm0 )
=(1.0×851.11×103×250/1.1)
= 193.43 kN-m
β= [(2Vu/Vd)-1)]2
=[(2×210×103/339.84×103)-1]2
= 0.055

4. Prof. Sudhir B. Gayake, SCOE Kopargaon
Md=Mdv =Md – β(Md –Mfd ) ≤ (1.2Zefy/ϒm0)
Md=Mdv =[ 193.43×106-0.055(193.43×106 – 144.75×106 )] ≤ (1.2×751.9×103 ×250/1.1)
=[190.75 kN-m] ≤ [205.06 kN-m] ………………………..(OK)
=[190.75 kN-m] > Mu=180kN-m……...………………….(OK)
5) Design Checks-
a) Check for web buckling= Fwb= B×tw×fcd
Assume stiff bearing length= b=150mm
B=(120+(h1/2))=(150+(288.3/2))= 294.15mm…………………….for end reaction
tw=7.4mm
fcd- depends on slenderness ratio (λ)=2.45(h1/tw)=95.45≈ 95
From table 9 (c) Pg.no. 42 and fy=250MPa
fcd= 114MPa
Fwb= B×tw×fcd =(294.15×7.4 ×114)= 248.14 kN>Vu=210kN………………(OK)

5. Prof. Sudhir B. Gayake, SCOE Kopargaon
b) Check for web crippling= Fw = Ae× (fyw/γm0)
fyw= 250MPa
Ae= b1×tw=(b+n1)×tw……………………………………………………for end reaction
=(150+ 2.5(tf+r1))× tw
=(150+2.5(11.4+16)) ×7.4
=1616.9mm2
Fw = Ae× (fyw/γm0)= (1616.9 ×250/1.1)=367.47kN>Vu=210kN………………….(OK)
Therefore ISLB 350 @ 49.5kg/m is safe to take given
moment and shear force
with check for buckling and crippling.